StringBuilder
- Kotlin course
- Beginner
- Lesson 91/116
- 5 min
Learn efficient string concatenation with StringBuilder.
StringBuilder: efficient concatenation
In Kotlin String is immutable; each + creates a new object. If you concatenate a lot in a loop, StringBuilder is more efficient.
val sb = StringBuilder()
sb.append("Hello")
sb.append(" ")
sb.append("world")
val result = sb.toString()
// "Hello world"StringBuilder grows the text in place; no new object per append. When you are done, call toString for the final string. With hundreds of appends the speed difference is clear.
StringBuilder vs String
String is immutable; + builds a new string every time. StringBuilder is mutable; append grows the same object.
var s = "a"
s = s + "b" // new String object
val sb = StringBuilder("a")
sb.append("b") // same objectWith hundreds of appends + gets slow; each step copies. StringBuilder avoids this cost. For a one-line concat + is fine.
Practice Concatenate three parts
Create a StringBuilder; append "Hello", ", " and "world" with append. Take the result with toString and print it.
Starter code
fun main() {
// Create a StringBuilder
// Append three parts
// Print the result with toString
}Sample solution
val sb = StringBuilder()
sb.append("Hello")
sb.append(", ")
sb.append("world")
println(sb.toString())Expected output
Merhaba, dünya
Questions and answers
Review the lesson questions together with their answers. To test yourself, solve the lesson interactively.
1 Match scenario to tool
| Term | Meaning |
|---|---|
One-shot concat |
String + |
Loop concat |
StringBuilder |
Final text |
toString |
Empty builder |
StringBuilder() |
2 Fill in the blank: declare StringBuilder
val sb = ____()val sb = StringBuilder()Explanation: StringBuilder() creates an empty builder.
3 You concatenate 1000 times in a loop. Which is faster?
- String and +
- Regex
- StringBuilder and append Correct answer
- matches
Explanation: StringBuilder does not create a new object per append.